Physics for Scientists and Engineers: A Strategic Approach 5th Edition · Work and Kinetic Energy · Problem 42
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Randall D. Knight — Work and Kinetic Energy: Problem 42
The energy used to pump liquids and gases through pipes is a significant fraction of the total energy consumption in the United States. Consider a small volume \(V\) of a liquid that has density \(\rho\). Assume that the fluid is nonviscous so that friction with the pipe walls can be neglected. a. An upward-pushing force from a pump lifts this volume of fluid a height \(h\) at constant speed. How much work does the pump do? b. The pump’s power is the rate at which it does work. The liquid’s volume flow rate \(Q\) is the volume of fluid per second, in \(m^3/s\), that enters or exits the pipe. Write an expression for the pump power \(P_{pump}\) needed to lift \(Q\) \(m^3/s\) of the liquid through height \(h\). This is a well-known and often-used equation in engineering. c. A 3-inch-diameter pipe can deliver 500 gpm (gallons per minute) of water. The viscosity of water is low and is often neglected in hydraulic calculations. In the United States, pumps are usually rated in horsepower (hp). What pump power in hp is needed to pump 500 gpm up a height of 3.0 m (\(\approx 10\) ft)? The density of water is \(1000 \text{ kg/m}^3\). Useful conversion factors are \(1 \text{ gal} = 3.79 \text{ L}\) and \(1 \text{ m}^3 = 1000 \text{ L}\).
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주어진 조건: . a, , in, . A, . In, 3.0 m
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Physics for Scientists and Engineers: A Strategic Approach · 5th Edition
저자: Randall D. Knight
출판사: Pearson
단원: Work and Kinetic Energy